无法合并链表？

Question

我正在尝试合并两个链表，但由于某种原因，当我的值被定义时，它显示无法读取null的值。

class Node {
  constructor(value) {
    this.value = value;
    this.next = null;
  }
}

class LinkedList {
  constructor() {
    this.head = null;
    this.length = 0;
  }

  add(value) {
    var node = new Node(value);
    var current;

    if (this.head == null) {
      this.head = node
    } else {
      current = this.head
      while (current.next) {
        current = current.next;
      }
      current.next = node
    }
    this.length++
  }
}

function mergeLinkedList(l1, l2) {
  let newList = new LinkedList()
  let currentNode1 = l1.head
  let currentNode2 = l2.head
  while (currentNode1 || currentNode2) {
    if (currentNode1.value <= currentNode2.value) {
      newList.add(currentNode1)
      newList.add(currentNode2)
    } else {
      newList.add(currentNode2)
      newList.add(currentNode1)
    }
    currentNode1 = currentNode1.next
    currentNode2.head = currentNode2.next
  }
  return newList
}

let l1 = new LinkedList()
l1.add(1)
l1.add(2)
l1.add(4)

let l2 = new LinkedList()
l2.add(1)
l2.add(3)
l2.add(4)

mergeLinkedList(l1, l2)

这应该返回一个新的链表，该链表的值是空的，但是在命中if语句时返回一个错误：if (currentNode1.value <= currentNode2.value)，声明它不能读取null的属性‘1->1->2->3->4->4’，但是这个值在这里定义得很清楚。

Answer 1

你有

while (currentNode1 || currentNode2)

因此，只需要在while循环中有一个节点不为null即可。但在那之后你就有了

if (currentNode1.value <= currentNode2.value)

它正在访问这两个节点的值。在这种情况下，您可能在while循环中使用currentNode2 == null，从而导致错误。

您应该将while循环条件编写为

 while (currentNode1 && currentNode2){
  .
  .
  .
 }

然后有另一个条件来检查您的链表，并附加一个还不为空的链表的其余部分。

Answer 2

您可以独立地检查节点并添加值，而不是节点本身。

​

class Node {
    constructor(value) {
        this.value = value;
        this.next = null;
    }
}

class LinkedList {
    constructor() {
        this.head = null;
        this.length = 0;
    }

    add(value) {
        var node = new Node(value);
        var current;

        if (this.head == null) {
            this.head = node
        } else {
            current = this.head
            while (current.next) {
                current = current.next;
            }
            current.next = node
        }
        this.length++
    }
}

function mergeLinkedList(l1, l2) {
    let newList = new LinkedList()
    let currentNode1 = l1.head
    let currentNode2 = l2.head
    while (currentNode1 && currentNode2) {
        if (currentNode1.value <= currentNode2.value) {
            newList.add(currentNode1.value)
            currentNode1 = currentNode1.next
        } else {
            newList.add(currentNode2.value)
            currentNode2 = currentNode2.next
        }
    }
    while (currentNode1) {
        newList.add(currentNode1.value)
        currentNode1 = currentNode1.next
    }
    while (currentNode2) {
        newList.add(currentNode2.value)
        currentNode2 = currentNode2.next
    }
    return newList
}

let l1 = new LinkedList()
l1.add(1)
l1.add(2)
l1.add(4)

console.log(l1);

let l2 = new LinkedList()
l2.add(1)
l2.add(3)
l2.add(4)

console.log(l2);

console.log(mergeLinkedList(l1, l2))

.as-console-wrapper { max-height: 100% !important; top: 0; }

​

Answer 3

​

​

通过查看上图：

var mergeTwoListss = function(l1, l2) {
    // initialize a dummy node, make a copy and iterate through with the copy. 
    // after iteration complete, you still have the reference to initial point
    const dummy=new ListNode()
    let tail= dummy
    // if you are comparing values in lists, make sure you have elements to compare 
    while (l1 && l2){
        if (l1.val<l2.val){
            tail.next=l1
            l1=l1.next
        }else{
            tail.next=l2
            l2=l2.next
        }
        tail=tail.next
    }
    // so far we have iterated for the same size of both nodes. what if one of them is longer. 
    if(l1){
        tail.next=l1
    }else if (l2){
        tail.next=l2
    }
    return dummy.next
    
};